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mamaluj [8]
2 years ago
14

Please help meeeeeeeeee ​

Mathematics
1 answer:
Yuliya22 [10]2 years ago
4 0

Answer:

FRACTION         DECIMAL                   PERCENT

3/100                   0.03                              3%

3/2                         1.5                                150%

13/20                    0.65                             65%

2/3                       0.6667                          66.67%

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What is the length of the hypotenuse? 24 42
AleksAgata [21]

Answer:

The length of the hypotenuse of a right triangle can be found using the Pythagorean theorem or a^2+b^2=c^2.

(24)^2+(42)^2=2340

Step-by-step explanation:

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Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company
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Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

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This means that n = 16

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that p = 0.5

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0

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P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018

P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085

Then

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

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